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This is incredibly relevant because $N(a_1,a_2,a_3)$ is a sum over polynomials of degree $4$ that are $3$-equivalent to $T^4+a_1T^3+a_2T^2+a_1T$ (and satisfying some factorization conditions). To make use of $G(j)$ we employ two orthogonality relations: $$(O1)\, q^{-j}\sum_{\chi \in G(j)} \chi(f_1)\overline{\chi(f_2)} = \mathbf{1}_{f_1 \text{ is }j\text{-equivalent to }f_2}$$ for any $f_1,f_2 \in \mathcal{M}_q$ and $$(O2)\, q^{-j}\sum_{f \in X_j} \chi_1(f)\overline{\chi_2(f)} = \mathbf{1}_{\chi_1=\chi_2} $$ where $\chi_1,\chi_2 \in G(j)$ and $X_j$ is a set of $q^j$ polynomials, none of which is $j$-equivalent to the other. Letting $$S = \{ f \in \mathcal{M}_q: \deg f = 4, \, f \text{ is a product of distinct linear factors over } \mathbb{F}_q \}$$ we can write $$N(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\overline{\chi( T^4+a_1T^3+a_2T^2+a_3T) }\sum_{f \in S} \chi(f) .$$ By $(O1)$ and $(O2)$, one obtains $$\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N^2(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2.$$ For the trivial character $\chi_0 \in G(3)$, $\sum_{f \in S} \chi_0(f) =|S|=\binom{q}{4}$. Moreover, $\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N(a_1,a_2,a_3) = |S| = \binom{q}{4}$. It follows that $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{1}{2} \left[q^{-3} \sum_{\chi_0\neq \chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2 + q^{-3}\binom{q}{4}^2 - \binom{q}{4}\right].$$ Using Newton's identities, one can write $$\sum_{f \in S} \chi(f) = \frac{1}{4!} \left( p_1^4 - 6p_1^2 p_2+3 p_2^2+8p_1p_3 - 6 p_4\right)$$ where $$p_k := \sum_{a \in \mathbb{F}_q} \chi^k(T+a).$$ By Weil's Riemann Hypothesis, each $p_k$ is $O(\sqrt{q})$ when $\chi\neq\chi_0$, which gives $\sum_{f \in S} \chi(f) = O(q^2)$, implying $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} =\frac{1}{2}q^{-3}\binom{q}{4}^2 + O(q^4) =\frac{q^5+O(q^4)}{1152}.$$ Relation to random matrix theory: Let us refine the above result. The total contribution of $\chi_0\neq \chi \in G(2)$ is $O(q^3)$ and we focus on $\chi \in G(3)\setminus G(2)$. By Newton's identities as above, analyzing $\sum_{f \in S} \chi(f)$ is the same as analyzing $(\sum_{a \in \mathbb{F}_q} \chi(T+a))^4$ (up to negligible error), which can be written as $(-q^{\frac{1}{2}} \mathrm{Tr}(\Theta_{\chi}))^4$ for some $\Theta_{\chi} \in U(2)$ ($\sum_{a \in \mathbb{F}_q} \chi(T+a)$ is just the linear term in an $L$-function $L(u,\chi)$ which has two zeros on $|u|=q^{-1/2}$). Using Theorem 1.2 of Katz, building on Deligne's equidistribution theorem, it follows that $$q^{-3} \sum_{\chi \in G(3)\setminus G(2)}\left| (q^{-\frac{1}{2}}\sum_{a \in \mathbb{F}_q} \chi(T+a))^4\right|^2 = \int_{U(2)} |\mathrm{Tr}(U)|^8{\rm d}U + O(q^{-1/2})=14+O(q^{-1/2})$$ and $(\star)$ follows after some algebra. The random matrix integral was evaluated using Theorem 1.1 of Rains.

This is incredibly relevant because $N(a_1,a_2,a_3)$ is a sum over polynomials of degree $4$ that are $3$-equivalent to $T^4+a_1T^3+a_2T^2+a_1T$ (and satisfying some factorization conditions). To make use of $G(j)$ we employ the orthogonality relation $$(O1)\, q^{-j}\sum_{\chi \in G(j)} \chi(f_1)\overline{\chi(f_2)} = \mathbf{1}_{f_1 \text{ is }j\text{-equivalent to }f_2}$$ for any $f_1,f_2 \in \mathcal{M}_q$. Letting $$S = \{ f \in \mathcal{M}_q: \deg f = 4, \, f \text{ is a product of distinct linear factors over } \mathbb{F}_q \}$$ we can write $$N(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\overline{\chi( T^4+a_1T^3+a_2T^2+a_3T) }\sum_{f \in S} \chi(f)$$ using $(O1)$ with $j=3$. Another relation is $$(O2)\, q^{-j}\sum_{f \in X_j} \chi_1(f)\overline{\chi_2(f)} = \mathbf{1}_{\chi_1=\chi_2} $$ where $\chi_1,\chi_2 \in G(j)$ and $X_j$ is a set of $q^j$ polynomials, none of which is $j$-equivalent to the other. From $(O1)$ and $(O2)$ (with $j=3$ and $X_3= \{ T^4+a_1 T^3+a_2 T^2+a_3T: (a_1,a_2,a_3) \in \mathbb{F}_q^3\}$) one obtains $$\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N^2(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2.$$ For the trivial character $\chi_0 \in G(3)$, $\sum_{f \in S} \chi_0(f) =|S|=\binom{q}{4}$. Moreover, $\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N(a_1,a_2,a_3) = |S| = \binom{q}{4}$. It follows that $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{1}{2} \left[q^{-3} \sum_{\chi_0\neq \chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2 + q^{-3}\binom{q}{4}^2 - \binom{q}{4}\right].$$ Using Newton's identities, one can write $$\sum_{f \in S} \chi(f) = \frac{1}{4!} \left( p_1^4 - 6p_1^2 p_2+3 p_2^2+8p_1p_3 - 6 p_4\right)$$ where $$p_k := \sum_{a \in \mathbb{F}_q} \chi^k(T+a).$$ By Weil's Riemann Hypothesis, each $p_k$ is $O(\sqrt{q})$ when $\chi\neq\chi_0$, which gives $\sum_{f \in S} \chi(f) = O(q^2)$, implying $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} =\frac{1}{2}q^{-3}\binom{q}{4}^2 + O(q^4) =\frac{q^5+O(q^4)}{1152}.$$ Relation to random matrix theory: Let us refine the above result. The total contribution of $\chi_0\neq \chi \in G(2)$ is $O(q^3)$ and we focus on $\chi \in G(3)\setminus G(2)$. By Newton's identities as above, analyzing $\sum_{f \in S} \chi(f)$ is the same as analyzing $(\sum_{a \in \mathbb{F}_q} \chi(T+a))^4$ (up to negligible error), which can be written as $(-q^{\frac{1}{2}} \mathrm{Tr}(\Theta_{\chi}))^4$ for some $\Theta_{\chi} \in U(2)$ ($\sum_{a \in \mathbb{F}_q} \chi(T+a)$ is just the linear term in an $L$-function $L(u,\chi)$ which has two zeros on $|u|=q^{-1/2}$). Using Theorem 1.2 of Katz, building on Deligne's equidistribution theorem, it follows that $$q^{-3} \sum_{\chi \in G(3)\setminus G(2)}\left| (q^{-\frac{1}{2}}\sum_{a \in \mathbb{F}_q} \chi(T+a))^4\right|^2 = \int_{U(2)} |\mathrm{Tr}(U)|^8{\rm d}U + O(q^{-1/2})=14+O(q^{-1/2})$$ and $(\star)$ follows after some algebra. The random matrix integral was evaluated using Theorem 1.1 of Rains.

This is incredibly relevant because $N(a_1,a_2,a_3)$ is a sum over polynomials of degree $4$ that are $3$-equivalent to $T^4+a_1T^3+a_2T^2+a_1T$ (and satisfying some factorization conditions). To mage use of $G(j)$ we employ two orthogonality relations: $$(O1)\, q^{-j}\sum_{\chi \in G(j)} \chi(f_1)\overline{\chi(f_2)} = \mathbf{1}_{f_1 \text{ is }j\text{-equivalent to }f_2}$$ for any $f_1,f_2 \in \mathcal{M}_q$ and $$(O2)\, q^{-j}\sum_{f \in X_j} \chi_1(f)\overline{\chi_2(f)} = \mathbf{1}_{\chi_1=\chi_2} $$ where $\chi_1,\chi_2 \in G(j)$ and $X_j$ is a set of $q^j$ polynomials, none of which is $j$-equivalent to the other. Letting $$S = \{ f \in \mathcal{M}_q: \deg f = 4, \, f \text{ is a product of distinct linear factors over } \mathbb{F}_q \}$$ we can write $$N(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\overline{\chi( T^4+a_1T^3+a_2T^2+a_3T) }\sum_{f \in S} \chi(f) .$$ By $(O1)$ and $(O2)$, one obtains $$\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N^2(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2.$$ For the trivial character $\chi_0 \in G(3)$, $\sum_{f \in S} \chi_0(f) =|S|=\binom{q}{4}$. Moreover, $\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N(a_1,a_2,a_3) = |S| = \binom{q}{4}$. It follows that $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{1}{2} \left[q^{-3} \sum_{\chi_0\neq \chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2 + q^{-3}\binom{q}{4}^2 - \binom{q}{4}\right].$$ Using Newton's identities, one can write $$\sum_{f \in S} \chi(f) = \frac{1}{4!} \left( p_1^4 - 6p_1^2 p_2+3 p_2^2+8p_1p_3 - 6 p_4\right)$$ where $$p_k := \sum_{a \in \mathbb{F}_q} \chi^k(T+a).$$ By Weil's Riemann Hypothesis, each $p_k$ is $O(\sqrt{q})$ when $\chi\neq\chi_0$, which gives $\sum_{f \in S} \chi(f) = O(q^2)$, implying $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} =\frac{1}{2}q^{-3}\binom{q}{4}^2 + O(q^4) =\frac{q^5+O(q^4)}{1152}.$$ Relation to random matrix theory: Let us refine the above result. The total contribution of $\chi_0\neq \chi \in G(2)$ is $O(q^3)$ and we focus on $\chi \in G(3)\setminus G(2)$. By Newton's identities as above, analyzing $\sum_{f \in S} \chi(f)$ is the same as analyzing $(\sum_{a \in \mathbb{F}_q} \chi(T+a))^4$ (up to negligible error), which can be written as $(-q^{\frac{1}{2}} \mathrm{Tr}(\Theta_{\chi}))^4$ for some $\Theta_{\chi} \in U(2)$ ($\sum_{a \in \mathbb{F}_q} \chi(T+a)$ is just the linear term in an $L$-function $L(u,\chi)$ which has two zeros on $|u|=q^{-1/2}$). Using Theorem 1.2 of Katz, building on Deligne's equidistribution theorem, it follows that $$q^{-3} \sum_{\chi \in G(3)\setminus G(2)}\left| (q^{-\frac{1}{2}}\sum_{a \in \mathbb{F}_q} \chi(T+a))^4\right|^2 = \int_{U(2)} |\mathrm{Tr}(U)|^8{\rm d}U + O(q^{-1/2})=14+O(q^{-1/2})$$ and $(\star)$ follows after some algebra. The random matrix integral was evaluated using Theorem 1.1 of Rains.

This is incredibly relevant because $N(a_1,a_2,a_3)$ is a sum over polynomials of degree $4$ that are $3$-equivalent to $T^4+a_1T^3+a_2T^2+a_1T$ (and satisfying some factorization conditions). To make use of $G(j)$ we employ two orthogonality relations: $$(O1)\, q^{-j}\sum_{\chi \in G(j)} \chi(f_1)\overline{\chi(f_2)} = \mathbf{1}_{f_1 \text{ is }j\text{-equivalent to }f_2}$$ for any $f_1,f_2 \in \mathcal{M}_q$ and $$(O2)\, q^{-j}\sum_{f \in X_j} \chi_1(f)\overline{\chi_2(f)} = \mathbf{1}_{\chi_1=\chi_2} $$ where $\chi_1,\chi_2 \in G(j)$ and $X_j$ is a set of $q^j$ polynomials, none of which is $j$-equivalent to the other. Letting $$S = \{ f \in \mathcal{M}_q: \deg f = 4, \, f \text{ is a product of distinct linear factors over } \mathbb{F}_q \}$$ we can write $$N(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\overline{\chi( T^4+a_1T^3+a_2T^2+a_3T) }\sum_{f \in S} \chi(f) .$$ By $(O1)$ and $(O2)$, one obtains $$\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N^2(a_1,a_2,a_3) = q^{-3} \sum_{\chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2.$$ For the trivial character $\chi_0 \in G(3)$, $\sum_{f \in S} \chi_0(f) =|S|=\binom{q}{4}$. Moreover, $\sum_{(a_1,a_2,a_3) \in \mathbb{F}_q^3} N(a_1,a_2,a_3) = |S| = \binom{q}{4}$. It follows that $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{1}{2} \left[q^{-3} \sum_{\chi_0\neq \chi \in G(3)}\left| \sum_{f \in S} \chi(f)\right|^2 + q^{-3}\binom{q}{4}^2 - \binom{q}{4}\right].$$ Using Newton's identities, one can write $$\sum_{f \in S} \chi(f) = \frac{1}{4!} \left( p_1^4 - 6p_1^2 p_2+3 p_2^2+8p_1p_3 - 6 p_4\right)$$ where $$p_k := \sum_{a \in \mathbb{F}_q} \chi^k(T+a).$$ By Weil's Riemann Hypothesis, each $p_k$ is $O(\sqrt{q})$ when $\chi\neq\chi_0$, which gives $\sum_{f \in S} \chi(f) = O(q^2)$, implying $$\sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} =\frac{1}{2}q^{-3}\binom{q}{4}^2 + O(q^4) =\frac{q^5+O(q^4)}{1152}.$$ Relation to random matrix theory: Let us refine the above result. The total contribution of $\chi_0\neq \chi \in G(2)$ is $O(q^3)$ and we focus on $\chi \in G(3)\setminus G(2)$. By Newton's identities as above, analyzing $\sum_{f \in S} \chi(f)$ is the same as analyzing $(\sum_{a \in \mathbb{F}_q} \chi(T+a))^4$ (up to negligible error), which can be written as $(-q^{\frac{1}{2}} \mathrm{Tr}(\Theta_{\chi}))^4$ for some $\Theta_{\chi} \in U(2)$ ($\sum_{a \in \mathbb{F}_q} \chi(T+a)$ is just the linear term in an $L$-function $L(u,\chi)$ which has two zeros on $|u|=q^{-1/2}$). Using Theorem 1.2 of Katz, building on Deligne's equidistribution theorem, it follows that $$q^{-3} \sum_{\chi \in G(3)\setminus G(2)}\left| (q^{-\frac{1}{2}}\sum_{a \in \mathbb{F}_q} \chi(T+a))^4\right|^2 = \int_{U(2)} |\mathrm{Tr}(U)|^8{\rm d}U + O(q^{-1/2})=14+O(q^{-1/2})$$ and $(\star)$ follows after some algebra. The random matrix integral was evaluated using Theorem 1.1 of Rains.

In addition to the illuminating answers of Will Sawin and Noam D. Elkies, I want to give a different perspective, coming from analytic number theory over $\mathbb{F}_q[T]$, and ultimately showing the question is equivalent to one about moments of cubic exponential sums. The computations below won't recover your formula in full, but they do show that $$(\star)\, \sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{q^5-22q^4+O(q^{7/2})}{1152}$$ holds when $(q,6)=1$. That is, they recover the first two terms in your formula. With more work this perspective should recover the full formula. The advantage of this perspective is that it is amenable to generalizations. From now on I assume that $(q,6)=1$ even if I do not say so explicitly.

Pointwise results: For any $a_1,a_2,a_3\in \mathbb{F}_q$, let $G$ be the Galois group of the polynomial $f(T)=T^4+a_1T^3+a_2T^2+a_3T+x$ over $\mathbb{F}_q(x)$. S. D. Cohen proved, in 1970, a general result that implies the following: $$N(a_1,a_2,a_3) = \frac{q}{|G|} + O(\sqrt{q})$$ holds as long as the splitting field of $T^4+a_1T^3+a_2T^2+a_3T+x$ does not contain $\mathbb{F}_{q^i}$ for $i>1$. Otherwise, $N(a_1,a_2,a_3)=O(\sqrt{q})$.

(If this looks funny, a different construction of such $\chi$-s proceeds by taking the even Dirichlet characters $\chi'$ modulo $T^{j+1}$ and then letting $\chi(f):=\chi'( f(1/T)T^{\deg f}/f(0))$ and $\chi(T)=1$. A 3rd construction is given at the end of this answer.)

In addition to the illuminating answers of Will Sawin and Noam D. Elkies, I want to give a different perspective, coming from analytic number theory over $\mathbb{F}_q[T]$, and ultimately showing the question is equivalent to one about moments of cubic exponential sums. The computations below currently only show that $$(\star)\, \sum_{(a_1,a_2,a_3)\in \mathbb{F}_q^3} \binom{N(a_1,a_2,a_3)}{2} = \frac{q^5-22q^4+O(q^{7/2})}{1152}$$ holds when $(q,6)=1$. That is, they recover the first two terms in your formula. With more work this perspective should recover the full formula. The advantage of this perspective is that it is amenable to generalizations. From now on I assume that $(q,6)=1$ even if I do not say so explicitly.

Pointwise results: For any $a_1,a_2,a_3\in \mathbb{F}_q$, let $G$ be the Galois group of the polynomial $f(T)=T^4+a_1T^3+a_2T^2+a_3T+x$ over $\mathbb{F}_q(x)$. S. D. Cohen proved, in 1970, a general result that implies the following: $$N(a_1,a_2,a_3) = \frac{q}{|G|} + O(\sqrt{q})$$ holds as long as the splitting field of $T^4+a_1T^3+a_2T^2+a_3T+x$ does not contain $\mathbb{F}_{q^i}$ for $i>1$. Otherwise, $N(a_1,a_2,a_3)=O(\sqrt{q})$. (In fact, this special case of Cohen's work seems to date to Birch and Swinnerton-Dyer.)

(If this looks funny, a different construction of such $\chi$-s proceeds by taking the even Dirichlet characters $\chi'$ modulo $T^{j+1}$ and then letting $\chi(f)=\chi'( f(1/T)T^{\deg f}/f(0))$ if $f(0)\neq 1$, and $\chi(T)=1$. A 3rd construction is given at the end of this answer.)